Chapter 28.

Direct-Current Circuits

Physics, 6th Edition

Chapter 28. Direct-Current Circuits
Resistors in Series and Parallel (Ignore internal resistances for batteries in this section.)
28-1. A 5-Ω resistor is connected in series with a 3-Ω resistor and a 16-V battery. What is the effective resistance and what is the current in the circuit? Re = R1 + R2 = 3 Ω +5 Ω; I= V 16 V = R 8Ω Re = 8.00 Ω I = 2.00 A 16 V 3Ω 5Ω

28-4. What is the equivalent resistance of 2, 4, and 6-Ω resistors connected in parallel?
2Ω 4Ω 6Ω

Re = 2 Ω + 4 Ω + 6 Ω;

Re = 12.0 Ω

129

Chapter 28. Direct-Current Circuits

Physics, 6th Edition

28-5. An 18-Ω resistor and a 9-Ω resistor are first connected in parallel and then in series with a 24-V battery. What is the effective resistance for each connection? Neglecting internal resistance, what is the total current delivered by the battery in each case? Re = R1 R2 (18 Ω)(9 Ω) = ; R1 + R2 18 Ω + 9 Ω I= V 24 V = ; R 6.00 Ω Re = 6.00 Ω
30 V 18 Ω 9Ω

I = 4.00 A Re = 27.0 Ω 18 Ω
24 V

Re = R1 + R2 = 18 Ω +9 Ω; I= V 24 V = R 27 Ω

9Ω

I = 0.889 A

28-6. A 12-Ω resistor and an 8-Ω resistor are first connected in parallel and then in series with a 28-V source of emf. What is the effective resistance and total current in each case? Re = R1 R2 (12 Ω)(8 Ω) = ; R1 + R2 12 Ω + 8 Ω I= V 28 V = ; R 4.80 Ω Re = 4.80 Ω 12 Ω
8Ω 28 V

28 V 12 Ω

8Ω

I = 5.83 A I= V 28 V = R 20 Ω

Re = R1 + R2 = 12 Ω +8 Ω;

Re = 20.0 Ω

I = 1.40 A

28-7. An 8-Ω resistor and a 3-Ω resistor are first connected in parallel and then in series with a 12-V source. Find the effective resistance and total current for each connection? Re = (3 Ω)(8 Ω) ; 3Ω+8Ω Re = 2.18 Ω I= V 12 V = ; R 2.18 Ω I= V 12 V = R 11 Ω I = 5.50 A

EMF and Terminal Potential Difference
28-16. A load resistance of 8 Ω is connected in series with a 18-V battery whose internal resistance is 1.0 Ω. What current is delivered and what is the terminal voltage? I= E 18 V = ; r + RL 1.0 Ω + 8 Ω I = 2.00 A

28-17. A resistance of 6 Ω is placed across a 12-V battery whose internal resistance is 0.3 Ω. What is the current delivered to the circuit? What is the terminal potential difference? I= E 12 V = ; r + RL 0.3 Ω + 6 Ω I = 1.90 A VT = 11.4 V

28-22. A 2-Ω and a 6-Ω resistor are connected in series with a 24-V battery of internal resistance 0.5 Ω. What is the terminal voltage and the power lost to internal resistance? Re = 2 Ω + 6 Ω + 0.5 Ω = 8.50 Ω; I= E 24 V = = 2.82 A Re 8.5 Ω

Kirchhoff’s Laws
28-25. Apply Kirchhoff’s second rule to the current loop in Fig. 28-22. What is the net voltage around the loop? What is the net IR drop? What is the current in the loop? Indicate output directions of emf’s, assume direction of current, and trace in a clockwise direction for loop rule:
2Ω 20 V

The minus sign means the current is counterclockwise (against the assume direction) *28-27. Use Kirchhoff’s laws to solve for the currents through the circuit shown as Fig. 28-23. First law at point P: I1 + I2 = I3 Current rule Loop A (2nd law): ΣE = ΣIR Loop rule
2 ΩP 4Ω I1

From (2): 1.5I1 + 3(0.439 A) = 3 A; and I1 = 0.536 A From (3): 3I2 – 5(0.439 A) = 0; and I2 = 0.736 A From (4): 5(0.439 A) + 6I4 = 6 A; and I4 = 0.634 A Currents in each branch are: I1 = 536 mA, I2 = 732 mA, I3 = 439 mA, I4 = 634 mA Note: Not all of the equations are independent. Elimination of two may yield another. It is best to start with the current rule, and use it to eliminate one of the currents quickly. 138

Chapter 28. Direct-Current Circuits

Physics, 6th Edition

The Wheatstone Bridge
28-30. A Wheatstone bridge is used to measure the resistance Rx of a coil of wire. The resistance box is adjusted for 6 Ω, and the contact key is positioned at the 45 cm mark when measured from point A of Fig. 28-13. Find Rx. Rx = R3l2 (6 Ω)(55 cm) = ; l1 (45 cm) ( Note: l1 + l2 = 100 cm ) Rx = 7.33 Ω

28-31. Commercially available Wheatstone bridges are portable and have a self-contained galvanometer. The ratio R2/R1 can be set at any integral power of ten between 0.001 and 1000 by a single dual switch. When this ratio is set to 100 and the known resistance R is adjusted to 46.7 Ω, the galvanometer current is zero. What is the unknown resistance? Rx = R3 R2 = (46.7 Ω)(100) ; R1 Rx = 4670 Ω

28-32. In a commercial Wheatstone bridge, R1 and R2 have the resistances of 20 and 40 Ω, respectively. If the resistance Rx is 14 Ω, what must be the known resistance R3 for zero galvanometer deflection? Rx = R3 R2 20 Ω = (14 Ω) ; R1 40 Ω Rx = 7.00 Ω

Challenge Problems:
28-33. Resistances of 3, 6, and 9 Ω are first connected in series and then in parallel with an 36-V source of potential difference. Neglecting internal resistance, what is the current leaving the positive terminal of the battery? Re = ΣRi = 3 Ω + 6 Ω + 9 Ω = 18 Ω ; I= 36 V ; 18 Ω I = 2.00 A

*28-35. Three resistors of 4, 8, and 12 Ω are connected in series with a battery. A switch allows the battery to be connected or disconnected from the circuit? When the switch is open, a voltmeter across the terminals of the battery reads 50 V. When the switch is closed, the voltmeter reads 48 V. What is the internal resistance in the battery? RL = 4 Ω + 8 Ω + 12 Ω = 24 Ω; I= 48 V = 2.00 A ; 24 V r= E = 50 V; VT = 48 V = IRL

E – VT = Ir; 50 V – 48 V = Ir

50 V - 48 V ; 2.00 A

r = 1.00 Ω

*28-36. The generator in Fig. 28-26 develops an emf of E1 = 24 V and has an internal resistance of 0.2 Ω. The generator is used to charge a battery E2 = 12 V whose internal resistance is 0.3 Ω. Assume that R1 = 4 Ω and R2 = 6 Ω. What is the terminal voltage across the generator? What is the terminal voltage across the battery? I= 24 V - 12 V = 1.14 A 6 Ω + 4 Ω + 0.2 Ω + 0.3 Ω 24 V r

*28-37. What is the power consumed in charging the battery for Problem 28-36. Show that the power delivered by the generator is equal to the power losses due to resistance and the power consumed in charging the battery. P = E I = (24 V)(1.143 A) Pe = 27.43 W 24 V r 4Ω 6Ω r 12 V

Critical Thinking Problems
*28-41. A three-way light bulb uses two resistors, a 50-W filament and a 100-W filament. A three-way switch allows each to be connected in series and provides a third possibility by connecting the two filaments in parallel? Draw a possible arrangement of switches than will accomplish these tasks. Assume that the household voltage is 120 V. What are the resistances of each filament? What is the power of the parallel combination? I1 Switch can be set at A, B, or C to give three possibilities: V2 (120 V) 2 (120 V) 2 P= ; R1 = = 288 Ω; R2 = = 144 Ω R 50 W 100 W For Parallel: Re = (288 Ω)(144 Ω) = 96 Ω 288 Ω + 144 Ω V 2 (120 V) 2 ; = R 96 Ω P = 150 W
R1 R2

B A

I2 C

IT
120 V

P=

142

Chapter 28. Direct-Current Circuits

Physics, 6th Edition

*28-42. The circuit illustrated in Fig. 28-7 consists of a 12-V battery, a 4-Ω resistor, and a switch. When new, the internal resistance of the battery is 0.4 Ω, and a voltmeter is placed across the terminals of the battery. What will be the reading of the voltmeter when the switch is open and when it is closed? After a long period of time, the experiment is repeated and it is noted that open circuit reading is unchanged, but the terminal voltage has reduced by 10 percent. How do you explain the lower terminal voltage? What is the internal resistance of the old battery? When new: I = E 12 V = ; R + r 4 Ω + 0.4 Ω I = 2.73 A VT = 10.9 V
4Ω